YES

We show the termination of the relative TRS R/S:

  R:
  f(|0|(),y) -> |0|()
  f(s(x),y) -> f(f(x,y),y)

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: f#(s(x),y) -> f#(f(x,y),y)
p2: f#(s(x),y) -> f#(x,y)

and R consists of:

r1: f(|0|(),y) -> |0|()
r2: f(s(x),y) -> f(f(x,y),y)
r3: rand(x) -> x
r4: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the non-minimal dependency pair problem (P, R), where P consists of

p1: f#(s(x),y) -> f#(f(x,y),y)
p2: f#(s(x),y) -> f#(x,y)

and R consists of:

r1: f(|0|(),y) -> |0|()
r2: f(s(x),y) -> f(f(x,y),y)
r3: rand(x) -> x
r4: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = x1 + ((1,0,0),(0,0,0),(0,1,0)) x2
      s_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (0,3,2)
      f_A(x1,x2) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (0,2,1)
      |0|_A() = (1,1,4)
      rand_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (1,1,0)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.